Added Solution2!

Week2/Solution2.ipynb

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+### 1 Exspectation Value and Variance of a Sample
+
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+1)
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+See lecture slide 46.
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+2)
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+$\left< V(x) \right> = \left< \frac{1}{N} \sum_{i} (x_{i}-\mu_{x})^{2} \right> = \frac{1}{N} \left<  \sum_{i} x_{i}^{2} - 2 x_{i} \mu_{x} + \mu_{x}^{2} \right> = \frac{1}{N}  \left( \sum_{i} \left< x_{i}^{2} \right> - 2 \left< x_{i} \right> \mu_{x} \right) + \mu_{x}^{2} = \left< x^{2} \right> - 2 \mu_{x}^{2} + \mu_{x}^{2} = \left< x^{2} \right> - \left< x \right>^{2} = \sigma_{x}^{2}$
+
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+By fixing a parameter for the measured mean like it was done in 1) one has to apply a correction factor. If the true mean is known this is not neccessary any more.
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+### 2 Poisson Distribution
+
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+$P_{k} (T) = \frac{1}{k!} e^{- \lambda T} (\lambda T)^{k}$
+
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+(lhs) $\dot{P}_{k}(T) = \frac{1}{k!} e^{- \lambda T} (- \lambda (\lambda T)^{k} + k \lambda^{k} T^{k-1} )$
+
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+(rhs) $\lambda [P_{k-1}(T) - P_{k}(T)] = \lambda \left[ \frac{e^{-\lambda T}}{(k-1)!} (\lambda T)^{k-1} - \frac{e^{-\lambda T}}{k!} (\lambda T)^{k} \right] = \frac{1}{k!} e^{- \lambda T} (- \lambda (\lambda T)^{k} + k \lambda^{k} T^{k-1} )$
+
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+$\Rightarrow \dot{P}_{k}(T) = \lambda [P_{k-1}(T) - P_{k}(T)]$
+
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+Differential Equation was derived by comparing small time intervall with a Taylor expansion, see lecture slide 62.
+
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+### 3 Bionomial- and Poisson Distribution
+
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+1)
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+Bionomial Distribtion for probobility for success with N trials!
+
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+$B(k; N, p) = \left(\begin{array}{c}
+N \\ k
+\end{array}\right) \cdot p^{k} \cdot (1-p)^{N-k}$
+
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+N = 60
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+p = 0.01
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+k = 0
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+$\Rightarrow B(0; 60, 0.01) = \left(\begin{array}{c}
+60 \\ 0
+\end{array}\right) \cdot 0.01^{0} \cdot (1-0.01)^{60-0} = 1 \cdot 1 \cdot (0.99)^{60} = 54.7 \%$
+
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+2)
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+Poisson Distribution for event which occur with an average event rate $\lambda$.
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+$P(k, \lambda) = \frac{\lambda^{k}}{k!} e^{- \lambda}$
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+$\lambda$ = $0.01 \cdot 60 = 0.6$
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+k = 0
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+$P(0,0.6) = \frac{0.6^{0}}{0!} e^{- 0.6} = 1 \cdot e^{-0.6} = 54.9 \%$
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+### 4 Poisson Distribution, Search for free Quarks
+
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+1)
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+$\sum_{i=1}^{110} P(i;229) = \sum_{i=1}^{110} \frac{229^{i}}{i!} e^{-229}= 1.6 \cdot 10^{-18} = p_{se}$
+
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+Very low propability, very unlikely to happen!
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+2)
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+N = 55000
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+$p_{me} = N \cdot p_{se} = 8.9 \cdot 10^{-14}$
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+3)
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+$\sum_{i=1}^{28} P(i;57) = \sum_{i=1}^{28} \frac{57^{i}}{i!} e^{-57}= 6.7 \cdot 10^{-6}$
+
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+Much more likely to happend!
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+4)
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+Combine both Poisson Distributions to achieve the new probability:
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+$\mu = 4$
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+$\mu \lambda = 229$
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+$\sum_{i=1}^{110} \sum_{N=0}^{\infty} P(i;N\mu) \cdot P(N;\lambda) = \sum_{i=1}^{110} \sum_{N=0}^{\infty} \frac{N \mu^{i}}{i!} e^{-N \mu} \cdot \frac{\lambda^{N}}{N!} e^{-\lambda} = 4.2 \cdot 10^{-5}$
+
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+Even more likely to happen.
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+### 5 Why are soccer results more random than handball results?
+
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+1)
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+$\lambda_{1} = 1$
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+$\lambda_{2} = 2$
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+$\sum_{i=1}^{\infty} \sum_{l=0}^{k-1} P(k;\lambda_{1}) \cdot P(l;\lambda_{2})= \sum_{i=1}^{\infty} \sum_{l=0}^{k-1}  \frac{\lambda_{2}^{k}}{k!} e^{-\lambda_{1}} \cdot \frac{\lambda_{2}^{l}}{l!} e^{-\lambda_{2}} = 18.3 \%$
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+2)
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+$\lambda_{1} = 10$
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+$\lambda_{2} = 20$
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+$\sum_{i=1}^{\infty} \sum_{l=0}^{k-1} P(k;\lambda_{1}) \cdot P(l;\lambda_{2})= 2.6 \%$
+
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+Summing up over all possible combinations, i=1 as at least one goal is needed to win a game.
+
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+### 5 Python Script (thanks to Joep Geuskens)
+
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+``` python
+from scipy.stats import poisson
+import numpy as np
+```
+
+%% Cell type:code id: tags:
+
+``` python
+def func(l=1):
+    arr  = np.array([poisson.cdf(k-1, 2*l)*poisson.pmf(k,l) for k in range(1,l*10)])
+    p    = np.sum(arr)
+    diff = arr[-1]/p          # relative size of the last term compared to the total probability
+    print(f"P(B>A)={p:.4f}")
+    print(f"Diff={diff:.2g}") # should be sufficiently small
+```
+
+%% Cell type:code id: tags:
+
+``` python
+print("1)")
+func(1)
+print("\n2)")
+func(10)
+```
+
+%%%% Output: stream
+
+    1)
+    P(B>A)=0.1826
+    Diff=5.6e-06
+    
+    2)
+    P(B>A)=0.0258
+    Diff=1.9e-60
+
+%% Cell type:markdown id: tags:
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+### 6 Multidimensional Gaussian
+
+%% Cell type:markdown id: tags:
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+$B =
+\frac{1}{\sigma_{1}^{2}\sigma_{2}^{2}- \rho^{2}\sigma_{1}^{2}\sigma_{2}^{2}}
+\left(\begin{array}{cc}
+\sigma_{2}^{2} & -c\\
+-c & \sigma_{1}^{2}
+\end{array}\right) =
+\frac{1}{\sigma_{1}^{2}\sigma_{2}^{2} \cdot (1- \rho^{2})}
+\left(\begin{array}{cc}
+\sigma_{2}^{2} & -c\\
+-c & \sigma_{1}^{2}
+\end{array}\right)$
+
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+$c = \rho x_{1} x_{2}$
+
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+$det B = \frac{1}{\sigma_{1}^{2}\sigma_{2}^{2} \cdot (1- \rho^{2})}$
+
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+$k =\sqrt{\frac{det B}{(2\pi)^{n}}} = \frac{1}{2 \pi \cdot \sigma_{1}\sigma_{2} \cdot \sqrt{(1- \rho^{2})}}$
+
+%% Cell type:markdown id: tags:
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+$$
+\left(\begin{array}{c}
+x_{1} \\ x_{2}
+\end{array}\right)
+\left(\begin{array}{cc}
+\sigma_{2}^{2} & -c\\
+-c & \sigma_{1}^{2}
+\end{array}\right)
+\left(\begin{array}{cc}
+x_{1} & x_{2}\\
+\end{array}\right) = x_{1}^{2}\sigma_{2}^{2} - 2 x_{1} x_{2} c + x_{2}^{2}\sigma_{1}^{2} =
+\sigma_{1}^{2} \sigma_{2}^{2} \left( \frac{x_{1}^2}{\sigma_{1}^{2}} + \frac{x_{2}^2}{\sigma_{2}^{2}} - 2 \rho \frac{x_{1}x_{2}}{\sigma_{1}\sigma_{2}} \right)
+$$
+
+%% Cell type:markdown id: tags:
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+$\Phi (x_{1}, x_{2}) = \frac{1}{2 \pi \cdot \sigma_{1}\sigma_{2} \cdot \sqrt{(1- \rho^{2})}} \exp \left( -\frac{1}{2 \cdot (1- \rho^{2})} \left( \frac{x_{1}^2}{\sigma_{1}^{2}} + \frac{x_{2}^2}{\sigma_{2}^{2}} - 2 \rho \frac{x_{1}x_{2}}{\sigma_{1}\sigma_{2}} \right) \right)$
+
+%% Cell type:markdown id: tags:
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+Substitue $x_{i}$ with $x_{i}- a_{i}$ to achieve the exspected form.
+
+%% Cell type:code id: tags:
+
+``` python
+```